Help please Protein produced from mutated strand? And effect of mutation??

2. 27.8 mL of an unknown were added to a 50.0-mL flask that weighs 464.7 g. The total mass of the flask and the liquid is 552.4

agasfer [191]

Answer:

$d=4.24x10^{-4}\frac{lb}{in^3}$

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:

$d=\frac{m}{V}$

Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:

$m=552.4g-464.7g=87.7g$

So that we are now able to calculate the density in g/mL first:

$d=\frac{87.7g}{27.8mL}=3.15g/mL$

Now, we proceed to the conversion to lb/in³ by using the following setup:

$d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}$

Regards!

6 0

3 years ago

Which is not a compound?Electrons spin in shells around the nucleus. The closest shell (n = 1) can contain a maximum of _____ el

melisa1 [442]

The closest shell (n = 1) can contain a maximum of 2 electrons.

6 0

3 years ago

If equal volumes of 0.1 M HCl and 0.2 M TRIS (base form) are mixed together. The pKa of TRIS is 8.30. Which of the following sta

blondinia [14]

Answer:

option D is correct

D. This solution is a good buffer.

Explanation:

TRIS (HOCH $_{2}$ ) $_{3}$ CNH $_{2}$

if TRIS is react with HCL it will form salt

(HOCH $_{2}$ ) $_{3}$ CNH $_{2}$ + HCL ⇆ (HOCH $_{2}$ ) $_{3}$ NH $_{3}$ CL

Let the reference volume is 100

Mole of TRIS is = 100 × 0.2 = 20

Mole of HCL is = 100 × 0.1 = 10

In the reaction all of the HCL will Consumed,10 moles of the salt will form

and 10 mole of TRIS will left

hence , Final product will be salt +TRIS(9 base)

H = Pk $_{a}$ + log (base/ acid)

8.3 + log(10/10)

8.3

6 0

3 years ago

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

3 years ago

100 POINTS!!! WILL GIVE BRAINLIEST FOR DETAILED AND CORRECT ANSWER!!!

sveticcg [70]

Answer:

dissolve rocksalt in heated water

Explanation:

If the rock salt is one large chunk, grind it into a powder using a mortar and pestle or a coffee grinder.

Add 30-50 milliliters of water to six heaping spatula scoops of rock salt.

Stir to dissolve the salt.

Place the filter paper in the mouth of the funnel.

Place the evaporating dish under the funnel to collect the liquid.

Slowly pour the rock salt solution into the funnel. Make sure you don't over-fill the funnel. You don't want the liquid to flow around the top of the filter paper because then it isn't getting filtered.

Save the liquid (filtrate) that comes through the filter. Many of the mineral contaminants did not dissolve in the water and were left behind on the filter paper.

Place the evaporating dish containing the filtrate on the tripod.

Position the Bunsen burner under the tripod.

Slowly and carefully heat the evaporating dish. Be careful! If you apply too much heat, you might break the dish.

Gently heat the filtrate until all the water is gone. It's okay if the salt crystals hiss and move a little.

Turn off the burner and collect your salt. Although some impurities will remain in the materials, many of them will have been removed simply by using the difference in solubility in water, mechanical filtration, and by applying heat to drive off volatile compounds.

If you want to further purify the salt, you can dissolve your product in hot water and crystallize the sodium chloride from it.

6 0

4 years ago